Two 13.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. T

Question

Two 13.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 13 V battery.

A) What is the charge on each electrode while the capacitor is attached to the battery? q1, q2 (need answers for both) The incorrect answer: 3.05 x 10^-10 C

B) What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.9 cm apart? The electrodes remain connected to the battery during this process. q1, q2 (need answers for both)

Explanation / Answer

A) q1 = C*v
where C = o*A/d = 8.854x10^-12**r^2/d = 8.854x10^-12**(0.065)^2/0.005 = 2.35 x10^-11F

So q1 = 2.35 x10^-11*13 = 3.05x10^-10C and q2 = -3.05x10^-10C

B)

(b)

New capacitance of the capacitor when they are moved apart is,'

C = o*A/d

C = o* pi*r^2 /d

C =8.85*10^-12 *pi * (0.13/2)^2 /1.9*10^-2

C =6.18 *10^ -12 F

The new charge on the capacitor plates is,

Q = CV =6.18 *10^ -12 F * 13 = 8.03*10^ -11 C

So the answer is ,

q1 = 8.03*10^ -11 C

q2 =- 8.03*10^ -11

Now C decreases to
C = o*A/d = 8.854x10^-12**r^2/d = 8.854x10^-12**(0.065)^2/0.019 = 6.18x10^-12F

q1 = 6.18x10^-12C and q2 = - 6.18x10^-12 Charge remains constant in this situation

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